Question: Subtract the following rational expressions. $\dfrac{4y^2}{7y-9}-\dfrac{8y^2}{y+12}=$
Answer: We can subtract two rational expressions whose denominators are equal by subtracting the numerators and keeping the denominator the same. [Does this fit with how we subtract rational numbers?] When the denominators are not the same, we must manipulate them so that they become the same. In other words, we must find a common denominator. Since the two denominators do not share any common factors, the common denominator is simply the product of these two denominators: $({7y-9})\cdot({y+12})$. Let's manipulate the expressions to have that denominator: $\begin{aligned} &\phantom{=}\dfrac{4y^2}{{7y-9}}-\dfrac{8y^2}{{y+12}} \\\\ &=\dfrac{4y^2\cdot({y+12})}{({7y-9})\cdot({y+12})}-\dfrac{8y^2\cdot({7y-9})}{({y+12})\cdot({7y-9})} \end{aligned}$ [Why did we do that?] Now that both denominators are the same, let's subtract! $\begin{aligned} &\phantom{=}\dfrac{4y^2\cdot(y+12)}{(7y-9)\cdot(y+12)}-\dfrac{8y^2\cdot(7y-9)}{(y+12)\cdot(7y-9)} \\\\ &=\dfrac{4y^2\cdot(y+12)-8y^2\cdot(7y-9)}{(7y-9)(y+12)} \\\\ &=\dfrac{4y^3+48y^2-56y^3+72y^2}{(7y-9)(y+12)} \\\\ &=\dfrac{-52y^3+120y^2}{(7y-9)(y+12)} \end{aligned}$ In conclusion, $\dfrac{4y^2}{7y-9}-\dfrac{8y^2}{y+12}=\dfrac{-52y^3+120y^2}{(7y-9)(y+12)}$